Refraction of Light and Total Internal Reflection

Light travels through a vacuum at $c = 3\times10^{8}\ \text{m/s}$ and the refractive index of water is $n = 1.33$.

1. a) Write down the relationship between the refractive index $n$, the speed of light in a vacuum $c$ and the speed of light in the medium $v$.
2. b) Calculate the speed of light in water.

A laser beam is incident from air ($n \approx 1$) onto the surface of water in an aquarium ($n = 1.33$) at an angle of incidence of $30^\circ$ to the normal. Given: $\sen 30^\circ = 0.5$. Observe the diagram:

1. a) Write down Snell's law of refraction, $n_1 \times \sen i = n_2 \times \sen r$, identifying each quantity.
2. b) Calculate the angle of refraction inside the water.
3. c) State whether the ray bends towards or away from the normal and why.

In the optics laboratory, a laser beam is incident from air ($n \approx 1$) onto a semicircular glass block of index $n = 1.52$ at an angle of incidence of $60^\circ$. Given: $\sen 60^\circ \approx 0.866$. Observe the diagram:

1. a) Calculate the angle of refraction in the glass.
2. b) Calculate the angle of deviation of the ray, that is, the difference between the angle of incidence and the angle of refraction.

A light source submerged in the water of a swimming pool ($n = 1.33$) emits a ray that reaches the surface at an angle of $20^\circ$ to the normal and emerges into the air ($n \approx 1$).

1. a) Without performing calculations, state whether the refracted ray bends towards or away from the normal.
2. b) Justify your answer using Snell's law, explaining what happens when the second medium has a lower refractive index.

A swimmer submerged in a swimming pool ($n_{\text{agua}} = 1.33$) looks up towards the surface. Air has $n \approx 1$. Observe the diagram:

1. a) Calculate the critical angle for the passage of light from water into air, that is, the angle of incidence for which the angle of refraction is $90^\circ$.
2. b) Explain what the swimmer observes when looking at points on the surface that form an angle with the vertical greater than that critical angle.

A block of glass of refractive index $n = 1.5$ is surrounded by air ($n \approx 1$).

1. a) Calculate the critical angle for the glass-air boundary.
2. b) A ray travelling through the glass reaches the surface at an angle of incidence of $30^\circ$: state, giving reasons, whether it emerges into the air or undergoes total internal reflection.
3. c) Answer the same question if the angle of incidence is $50^\circ$.

In an optical fiber, light travels through the core without escaping to the outside even if the fiber bends. Observe the diagram:

1. a) Which optical phenomenon keeps the light inside the core?
2. b) What condition must the angle of incidence of the ray on the core wall satisfy for this to occur?
3. c) Must the refractive index of the core be greater or smaller than that of the surrounding cladding? Give reasons for your answer.

An optical fiber has a core of index $n_1 = 1.48$ and a cladding of index $n_2 = 1.46$.

1. a) Calculate the critical angle at the core-cladding boundary.
2. b) A ray in the core reaches that boundary with an angle of incidence of $85^\circ$ with respect to the normal: state, with justification, whether it is transmitted into the cladding or continues to be guided along the core.

An optical fiber with a core index of $n_1 = 1.50$ and a cladding index of $n_2 = 1.47$ is in air ($n \approx 1$). It can be shown that the maximum angle of entry $\theta_{\text{max}}$ for a ray to be guided by the fiber satisfies $\sen\theta_{\text{max}} = \sqrt{n_1^2 - n_2^2}$. Observe the diagram:

1. a) Calculate the critical angle at the core-cladding interface.
2. b) Calculate the maximum acceptance angle $\theta_{\text{max}}$ of the fiber.
3. c) Interpret the result: explain what the acceptance cone is and why it matters when launching light into the fiber, for example in a medical endoscope.

A ray of light enters a glass plate of refractive index $n = 1.5$ from air ($n \approx 1$) at an angle of incidence of $45^\circ$. Given: $\sen 45^\circ \approx 0.71$.

1. a) Calculate the angle of refraction by applying Snell's law.
2. b) Draw in your notebook the boundary surface, the normal, the incident ray and the refracted ray, labelling both angles.

You can do it on paper and send a photo with your answer.

A ray of light is incident at $45^\circ$ on a glass sheet with flat, parallel faces, with refractive index $n = 1.5$ and thickness $e = 2\ \text{cm}$, surrounded by air ($n \approx 1$). Data: $\sen 45^\circ \approx 0.707$; once you have calculated the angle of refraction $r$, you can use $\cos r \approx 0.88$ and $\sen(45^\circ - r) \approx 0.29$. Observe the diagram:

1. a) Calculate the angle of refraction $r$ at the first face.
2. b) Applying Snell's law at the second face, show that the emergent ray leaves at $45^\circ$, that is, parallel to the incident ray.
3. c) The lateral displacement of the ray is $d = e \times \sen(i - r)/\cos r$. Calculate its value and explain its meaning with a drawing in your notebook.

A coin rests at the bottom of a swimming pool with a depth of $2\ \text{m}$ ($n_{\text{agua}} = 1.33$, $n_{\text{aire}} \approx 1$). For an observer looking almost vertically downwards, the angles are small and the approximation $\sen\theta \approx \tan\theta$ can be used. Observe the diagram:

1. a) Using Snell's law and this approximation, deduce that the apparent depth is $h' = h/n$.
2. b) Calculate the apparent depth of the coin.
3. c) Explain why the pool appears shallower than it actually is and what practical risk this perception poses.